Engineering KVL and finding circuit currents

[cjm181]

Homework Statement

The diagram shown below shows a bi-directional opto coupler input
interface circuit. When a supply voltage of 20 V is applied the LED
carries a current and 2 V is dropped across it. Calculate the value of
the LED current and the value of current through the 3 kΩ resistance.

Homework Equations

The Attempt at a Solution

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As the voltage in a parallel circuit is the same in each leg, for a voltage drop of 2V across the LED, the 470Ohm resistor must also have a voltage drop of 2V.

So the voltage drop across the 3kohm resistor using KVL is:
E - V1 - V2 = 0
V1 = E - V2
V1 = 20 - 2 = 18V

So I now have V and R for the 3k resistor, so using Ohms Law:
I = V / R
I = 18 / 3000 = 6x10^-3A or 6mA
6mA is the current flowing through the 3k resistor

As the 470 resistor has a voltage drop of 2V, I have V and I

I = V / R
I1 = 2 / 470 = 4.255x10^-3A = 4.255mA

Current in a parallel circuit is the sum of the current in each leg, so

I = I1 + I2

So
I2 = I - I1
I2 = 6 - 4.255 = 1.745mA

The answers are the current flowing through the LED is 1.745mA and the current flowing through the 3k resistor is 6mA

Is this looking somewhere close. I struggled with this

Kr
Craig

 

[gneill]

Your method and result values look good. You might want to round the final results to appropriate numbers of significant figures.